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Ask iit jee aieee pet cbse icse state board experts Expert Question: CIRCULAR MOTION
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[Post New]7 Apr 2008 14:39:47 IST
    Subject: CIRCULAR MOTION
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varun.tinkle (1160)

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A HEMISPHERICAL BOWL OF RADIUS R IS ROTATED ABT ITS AXIS OF SYMMETRY WHICH IS KEPT VERTICAL.A SMALL BLOCK IS KEPTON THE BOWL IN THE POSITION WHERE THE RADIUS MAKES AN ANGLE@ WITH THE VERTICAL.THE BLOCK ROTATES IN THE BOWL WITHOUTH SLIPPING.THE FRICTION COEFFICIENT BETWEEN THE BLOCK AND THE BOWL IS u.FIND THE RANGE OF ANGULAR SPEED FOR WHICH THE BLOCK WILL NOT SLIP.
PLS SOLVE IT IN DETAIL
PLS EXPLN IN DETAI AT WHICH ANGLE AND DIRECTION WILL THE FRICTIONAL FORCE ACT
From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.

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[Post New]7 Apr 2008 15:36:31 IST
    Subject: Re:CIRCULAR MOTION
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sboosy (3063)

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\mbox{The block is going in a circle of radius} \ R\sin(\alpha) \\ \\ \mbox{The forces concerned are mg ,N , Friction} \\ \\ \mbox{First consider the case where w is such that block tries to slip down} \\ \\ \mbox{Here friction acts up} \\ \\ \mbox{Thus splitting into components we have} \\ \\ N\sin(\alpha) - \mu N\cos(\alpha) = m(R\sin(\alpha))w^2 \\ \\ N\cos(\alpha) + \mu N\sin(\alpha) = mg \\ \\ \mbox{Dividing we get} \\ \\ w =\sqrt{\frac{g(\sin(\alpha) - \mu\cos(\alpha))}{R\sin(\alpha)(\cos(\alpha) + \mu\sin(\alpha))}} ....(1) \\ \\ \mbox{Now taking the other case we have} \\ \\ N\sin(\alpha) + \mu N\cos(\alpha) = m(R\sin(\alpha))w^2 \\ \\ N\cos(\alpha) - \mu N\sin(\alpha) = mg \\ \\ \mbox{In this case} \\ \\ w=\sqrt{\frac{g(\sin(\alpha) + \mu\cos(\alpha))}{R\sin(\alpha)(\cos(\alpha) - \mu\sin(\alpha))}} .....(2) \\ \\ \mbox{Thus w lies between 1 and 2}
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[Post New]7 Apr 2008 16:49:56 IST
    Subject: CIRCULAR MOTION
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varun.tinkle (1160)

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AT WHAT ANGLE WILL THE FRICTIONAL FORCE ACT
From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.

MY BLOG
http://my.opera.com/vu2rje/blog/
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[Post New]7 Apr 2008 17:13:00 IST
    Subject: CIRCULAR MOTION
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varun.tinkle (1160)

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PLS DRAW THE DIAGRAM IF U CAN
From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.

MY BLOG
http://my.opera.com/vu2rje/blog/
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[Post New]7 Apr 2008 17:17:25 IST
    Subject: Re:CIRCULAR MOTION
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arpan1 (665)

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check this
http://www.goiit.com/posts/list/inorganic-chemistry-ridiculous-51704.htm#259119
all the best ...
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[Post New]7 Apr 2008 17:18:48 IST
    Subject: CIRCULAR MOTION
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varun.tinkle (1160)

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HEY DUDE PLS IM TRYING TO GET AN ANSWER
From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.

MY BLOG
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[Post New]7 Apr 2008 17:33:19 IST
    Subject: CIRCULAR MOTION
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akhil_o (2709)

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The normal force is towards the centre of circle
the frictional force is perpendicular to the normal ie towards tangent to the circle at that point
" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates
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[Post New]7 Apr 2008 17:41:04 IST
    Subject: CIRCULAR MOTION
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varun.tinkle (1160)

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why
From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.

MY BLOG
http://my.opera.com/vu2rje/blog/
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[Post New]7 Apr 2008 19:29:32 IST
    Subject: Re:CIRCULAR MOTION
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varun.tinkle (1160)

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BAL GANESH GAVE ME THIS DIAGRQAM
CAN ANYONE EXPLN WHY DOES FRICTION ACT IN THIS DIRECTION

From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.

MY BLOG
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[Post New]7 Apr 2008 19:30:42 IST
    Subject: Re:CIRCULAR MOTION
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varun.tinkle (1160)

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Re:CIRCULAR MOTION

From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.

MY BLOG
http://my.opera.com/vu2rje/blog/
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[Post New]7 Apr 2008 20:59:03 IST
    Subject: CIRCULAR MOTION
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akhil_o (2709)

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In the circle path,
the path of the particle is restricted to the circle path, and the velocity has to be perpendicular to the normal, ie the tangent
the frictional force opposes this motion hence it is also in the same direction, tangential but opposite in direction
" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
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[Post New]8 Apr 2008 14:10:17 IST
    Subject: Re:CIRCULAR MOTION
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varun.tinkle (1160)

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just clear one thing
the resultant force on the particle is the resultant of the tangential force and centripital force
but since the centripital force is due to the particle it can be left out
am i right
if i am then the angle between the normal force and the vertical is @
and between the frictional force and the horizontal force is @
if so then could u prove it pls
From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.

MY BLOG
http://my.opera.com/vu2rje/blog/
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[Post New]8 Apr 2008 14:44:03 IST
    Subject: Re:CIRCULAR MOTION
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anandghegde (1712)

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Centripetal force is not just another force which you can add in your FBD.........the resultant of all other forces provide the required centripetal force.....
"I a universe of atoms.......an atom in the universe"
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[Post New]8 Apr 2008 16:18:45 IST
    Subject: CIRCULAR MOTION
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varun.tinkle (1160)

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so that means the motion of the particle is along the tangent
From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.

MY BLOG
http://my.opera.com/vu2rje/blog/
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[Post New]8 Apr 2008 16:47:19 IST
    Subject: Re:CIRCULAR MOTION
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